Image of line (x - 2/3)=(y - 1/1)=(z - 1/- 4) in the plane x+y+z=7 is

Question

Image of line (x - 2/3)=(y - 1/1)=(z - 1/- 4) in the plane x+y+z=7 is (A) (x - 4/1)=(y - 3/1)=(z - 3/1) (B) (x - 3/1)=(y - 4/1)=(z - 3/1) (C) (x - 4/3)=(y - 3/1)=(z - 3/- 4) (D) (x - 3/3)=(y - 4/1)=(z - 3/- 4)

Tardigrade Admin · Accepted Answer

The line is perpendicular to normal, hence the line is parallel to plane, so the reflected line has the same direction.
Now, image of

(2, 1, 1)

in the plane is

(α, β, γ)

say, then

\frac{α - 2}{1} = \frac{β - 1}{1} = \frac{γ - 1}{1} = \frac{- 2 ( 4 - 7 )}{1 + 1 + 1} = 2

So, the point is

(4, 3, 3)

Q. Image of line $\frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 1}{- 4}$ in the plane $x + y + z = 7$ is

$\frac{x - 4}{1} = \frac{y - 3}{1} = \frac{z - 3}{1}$

$\frac{x - 3}{1} = \frac{y - 4}{1} = \frac{z - 3}{1}$

$\frac{x - 4}{3} = \frac{y - 3}{1} = \frac{z - 3}{- 4}$

$\frac{x - 3}{3} = \frac{y - 4}{1} = \frac{z - 3}{- 4}$

Q. Image of line 3x−2​=1y−1​=−4z−1​ in the plane x+y+z=7 is

1x−4​=1y−3​=1z−3​

1x−3​=1y−4​=1z−3​

3x−4​=1y−3​=−4z−3​

3x−3​=1y−4​=−4z−3​

The line is perpendicular to normal, hence the line is parallel to plane, so the reflected line has the same direction. Now, image of (2,1,1) in the plane is (α,β,γ) say, then 1α−2​=1β−1​=1γ−1​=1+1+1−2(4−7)​=2 So, the point is (4,3,3)

Q. Image of line $\frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 1}{- 4}$ in the plane $x + y + z = 7$ is

$\frac{x - 4}{1} = \frac{y - 3}{1} = \frac{z - 3}{1}$

$\frac{x - 3}{1} = \frac{y - 4}{1} = \frac{z - 3}{1}$

$\frac{x - 4}{3} = \frac{y - 3}{1} = \frac{z - 3}{- 4}$

$\frac{x - 3}{3} = \frac{y - 4}{1} = \frac{z - 3}{- 4}$